SSC (Marathi Semi-English) 10thMaharashtra State Board
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Concept: Solutions of Quadratic Equations by Factorization

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Notes

By substituting arbitrary values for the variable and deciding the roots of quadratic equation is a time consuming process. Let us learn to use factorisation method to find the roots of the given quadratic equation.
        x2 - 4 x - 5 = (x - 5) (x + 1)
(x - 5) and (x + 1) are two linear factors of quadratic polynomial x2 - 4 x - 5.
So the quadratic equation obtained from x2 - 4 x - 5 can be written as (x - 5) (x + 1) = 0 

If product of two numbers is zero then at least one of them is zero.
     x - 5 = 0 or x + 1 = 0
  ∴x = 5 or x = -1
  ∴5 and the -1 are the roots of the given quadratic equation.
While solving the equation first we obtained the linear factors. So we call this method as ’factorization method’ of solving quadratic equation.

Ex.(1)  m2 - 14 m + 13 = 0

∴m2 - 13 m - 1m + 13 = 0
∴m (m - 13) -1 (m - 13) = 0
∴(m - 13) (m - 1) = 0
∴m - 13 = 0 or m - 1 = 0
∴m = 13 or m = 1

∴13 and 1 are the roots of the given quadratic equation.

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